Question

A parallel plate capacitor of plate area A and separation $d$ is charged to a potential difference and then the battery is disconnected. A slab of dielectric constant $K$ is then inserted between the plates of the capacitor so as to fill the whole space between the plates. Find the work done on the system and the process of inserting the slab.
(A) $\dfrac{{\varepsilon A{V^2}}}{{2d}}\left[ {1 - \dfrac{1}{K}} \right]$
(B) $\dfrac{{\varepsilon A{V^2}}}{d}\left[ {\dfrac{1}{K} - 1} \right]$
(C) $\dfrac{{\varepsilon A{V^2}}}{{2d}}\left[ {\dfrac{1}{K} + 1} \right]$
(D) $W = \dfrac{{\varepsilon A{V^2}}}{d}\left[ {\dfrac{1}{K} + 1} \right]$

Answer 1

Hint : The charge of a disconnected capacitor remains the same before and after the dielectric has been placed. The voltage drops during after the dielectric has been placed.
Formula used: In this solution we will be using the following formulae;
$Q = CV$where $Q$ is the charge on the capacitor, $C$ is the capacitance of the capacitor and $V$ is the voltage across its plate.
$C = \dfrac{{K\varepsilon A}}{d}$ where $K$ is the dielectric constant for the material between the plate, $\varepsilon$ is the permittivity of free space, $A$ is the area of the capacitor plates, and $d$ is the distance between the plates.
$U = \dfrac{1}{2}C{V^2}$ where $U$ is the potential energy (or energy) possessed by a capacitor.
$W = - \Delta U$ where $W$ is the work done, and $\Delta$ signifies change in a quantity (in this case, $U$)

Complete Step-by-Step solution:
To solve, we note that the charge before and after the dielectric has been placed are the same, since the capacitor was disconnected before it was done.
Generally,
$Q = CV$ where $Q$ is the charge on the capacitor, $C$ is the capacitance of the capacitor and $V$ is the voltage across its plate.
Capacitance is
$C = \dfrac{{K\varepsilon A}}{d}$ where $K$ is the dielectric constant for the material between the plate, $\varepsilon$ is the permittivity of free space, $A$ is the area of the capacitor plates, and $d$ is the distance between the plates.
The initial capacitance is
${C_0} = \dfrac{{\varepsilon A}}{d}$ (since $K = 1$ for air)
Hence, charge is,
$Q = \dfrac{{\varepsilon A}}{d}V$
The final capacitance
$C = \dfrac{{K\varepsilon A}}{d}$
Hence, charge is also
$Q = \dfrac{{K\varepsilon A}}{d}{V_f}$
Hence, by equating, we have
$Q = \dfrac{{K\varepsilon A}}{d}{V_f} = \dfrac{{\varepsilon A}}{d}V$
Then, by simplification, we have
${V_f} = \dfrac{V}{K}$.
Now the potential energy is given as
$U = \dfrac{1}{2}C{V^2}$ where $U$ is the potential energy (or energy) possessed by a capacitor.
Hence, the initial and final energy are respectively
${U_i} = \dfrac{1}{2}\left( {\dfrac{{\varepsilon A}}{d}} \right){V^2}$
${U_f} = \dfrac{1}{2}\left( {\dfrac{{K\varepsilon A}}{d}} \right)V_f^2 = \dfrac{1}{2}\left( {\dfrac{{K\varepsilon A}}{d}} \right)\dfrac{{{V^2}}}{{{K^2}}} = \dfrac{1}{2}\left( {\dfrac{{\varepsilon A}}{d}} \right)\dfrac{{{V^2}}}{K}$
The work done is
$W = - \Delta U$ where $W$ is the work done, and $\Delta$ signifies change in a quantity (in this case, $U$)
Hence,
$W = - \left[ {\dfrac{1}{2}\left( {\dfrac{{\varepsilon A}}{d}} \right)\dfrac{{{V^2}}}{K} - \dfrac{1}{2}\left( {\dfrac{{\varepsilon A}}{d}} \right){V^2}} \right]$
$\Rightarrow W = - \dfrac{1}{2}\left( {\dfrac{{\varepsilon A}}{d}} \right){V^2}\left[ {\dfrac{1}{K} - 1} \right]$
By rearranging, we have
$W = \dfrac{{\varepsilon A}}{{2d}}{V^2}\left[ {1 - \dfrac{1}{K}} \right]$

Hence, the correct option is A

Note: Alternatively, without lengthy calculations and using some reasoning, we can get the answer. We can reason as follows: first work done is a difference between different energy states, hence the answer cannot be C or D, also, energy is $\dfrac{1}{2}C{V^2}$, hence, the answer cannot be B either. Hence, the answer is A.